Estimating $e^{0.99}$ using a Maclaurin polynomial, what is the least degree of the polynomial that assures an error smaller than $0.001$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $3$ (Choice B) B $4$ (Choice C) C $5$ (Choice D) D $6$
Solution: We will use the Lagrange error bound. Let's assume the polynomial's degree is $n$. The $(n+1)^{\text{th}}$ derivative of $e^x$ is $e^x$. On the interval between $x=0$ and $x=0.99$, the greatest value of the derivative is $e^{0.99}\approx2.7$. The Lagrange bound for the error assures that $|R_n(0.99)|\leq \dfrac{2.7}{(n+1)!}(0.99)^{n+1}$. Solving $\dfrac{2.7}{(n+1)!}0.99^{n+1}<0.001$ using trial and error, we find that $n\geq6$. In conclusion, the least degree of the polynomial that assures our error bound is $6$.